F.4 Maths Joint Variation

It is known that y varies inversely as x² and √z where z > 0, y = 5 when x = 2 and z = 25.

(a) Express y in terms of x and z.
y ∝ 1/(x²√z)
y = k/(x²√z) where k ≠ 0 is a constant.
It is given that y = 5 when x = 2 and z = 25.
That is, 5 = k/(2²√25) = k/20 ⇒ k = 100.
Thus, y = 100/(x²√z).

(b) Find the value of x when y = 3 and z = 144.
y = 100/(x²√z)
3 = 100/(x²√144)
3 = 100/(12x²)
36x² = 100
x² = 100/36 = 25/9
x = ±5/3

2014-02-24 13:38:13 補充：
郭老師午安～

ღ(｡◕‿◠｡)ღ

早晨！好快呀！

~~~~~~~~~~~~~~~